Question 1149653: During a period of 11 years, of 1800 that were selected for grand jury duty, 25% of them were immigrants. Construct a 99% confidence interval for the percentage of grand jury members who were immigrants.
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
Given information
The value "11 years" is not used at all.
phat = 0.25 is the sample proportion
n = 1800 is the sample size
At 99% confidence, the critical z value is approximately z = 2.576
I used this table to find the critical z value. A similar table should be found in the appendix section of your textbook.
If you choose to use the table I have linked, then locate the bottom row that shows "99% confidence level". Afterward look just above that to find 2.576
Alternatively you can use the invNorm function on your TI83 or TI84 calculator to get the same approximate value.
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Lets compute the lower bound (L) and the upper bound (U) of the confidence interval.
L = lower bound of confidence interval
L = phat - z*sqrt(phat*(1-phat)/n)
L = 0.25 - 2.576*sqrt(0.25*(1-0.25)/1800)
L = 0.22370881009412
L = 0.22
U = upper bound of confidence interval
U = phat + z*sqrt(phat*(1-phat)/n)
U = 0.25 + 2.576*sqrt(0.25*(1-0.25)/1800)
U = 0.27629118990588
U = 0.28
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The 99% confidence interval is (0.22, 0.28)
This means at 99% confidence we can say 0.22 < p < 0.28, where p is the population proportion of immigrants selected for grand jury duty.
This is over the time span of the 11 years mentioned.
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